MHB Particle Velocity: Find $|v(2)|$ & $s(2)$

[karush]

https://www.physicsforums.com/attachments/5144

$a(t)=3-2t$

$v\left(t\right)=\int a\left(t\right) dt = - t^2 +3t+3$

$\left| v\left(2\right) \right|=5$

$s\left(t\right)=\int v\left(t \right) dt $

$s\left(2\right)=9\frac{1}{3}$

 

[MarkFL]

I would begin with:

$$\d{v}{t}=3-2t$$

Integrate w.r.t $t$:

$$\int_3^{v(t)}\,du=\int_0^t 3-2w\,dw$$

Apply the FTOC and solve for $v(t)$:

$$v(t)=-t^2+3t+3$$

They have asked for the velocity, which is a vector, so you want both the magnitude and the sign (which signifies direction for straight-line motion).

$$v(2)=-(2)^2+3(2)+3=5$$

Next, find the position:

$$\d{x}{t}=-t^2+3t+3$$

Integrate w.r.t $t$:

$$\int_0^{x(t)}\,du=\int_0^t -w^2+3w+3\,dw$$

Apply the FTOC:

$$x(t)=-\frac{1}{3}t^3+\frac{3}{2}t^2+3t$$

Now, distance is the magnitude of the position vector, so we should use the absolute value of the position to get the distance.

Hence:

$$|x(2)|=\left|-\frac{1}{3}(2)^3+\frac{3}{2}(2)^2+3(2)\right|=\left|-\frac{8}{3}+6+6\right|=\frac{28}{3}$$

So yes, I agree with your results. :)

 

[karush]

That was very helpful but what is FTOC?

I will post some more of these it's still a foggy topic

 

[MarkFL]

That was very helpful but what is FTOC?

I will post some more of these it's still a foggy topic

I was speaking of the anti-derivative form of the FTOC (Fundamental Theorem Of Calculus), for definite integrals:

$$\int_a^b f(x)\,dx=F(b)-F(a)$$ where $$\d{F}{x}=f(x)$$.